$ f(x)=\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{{{x}^{2n+3}}}{\left( 2n+1 \right)!}}$ $f'''(0)=$
Note that because we are interested in what happens at $x=0$, we only need to pay attention to the constant term in the series that represents $f'''$. With that said, we now calculate derivatives. For clarity, we will work out one more term than we actually need to do the problem. $\begin{aligned} f'(x)&=3x^2-\dfrac{5{{x}^{4}}}{3!}+... \\\\ f''(x)&=6x-\dfrac{20{{x}^{3}}}{3!}+... \\\\ f'''(x)&=6-\dfrac{60x^2}{3!}+... \end{aligned}$ So $f'''(0)=6$.